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3 years ago

Which of the following is a galvanic cell?

Chemistry

2 answers:

Dafna11 [192]3 years ago

8 0

Answer:

C. a Pe X Apporved

Explanation:

Trust

sladkih [1.3K]3 years ago

7 0

C. Aluminum (Al) oxidized, zinc (Zn) reduced

<h3>Further explanation</h3>

Given

Metals that undergo oxidation and reduction

Required

A galvanic cell

Solution

The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.

$\large {\boxed {\bold {E ^ ocell = E ^ ocatode -E ^ oanode}}}$

or:

E ° cell = E ° reduction-E ° oxidation

For the reaction to occur spontaneously (so that it E cell is positive), the E° anode must be less than the E°cathode

If we look at the voltaic series:

The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)

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From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.

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Answer:

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Explanation:

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The overall reaction equation, and the rate expression. The two;

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3 years ago

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Answer:

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2 years ago

Read 2 more answers

What will be the final temperature of the solution in a coffee cup calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is added

padilas [110]

Answer:

21.21°C will be the final temperature of the solution in a coffee cup calorimeter.

Explanation:

$HCl+NaOH\rightarrow H_2O+NaCl$

$\Delta H$ = enthalpy change = -57.2 kJ/mol of NaOH

Moles of sodium hydroxide = n

Molarity of the NaOH = 0.250 M

Volume of NaOH solution = V = 50.00 mL = 0.050 L

$n=Molarity\times V=0.250 M\times 0.050 L= 0.0125 mol$

Moles of HCl = n'

Molarity of the HCl= 0.250 M

Volume of HCl solution = V' = 50.00 mL = 0.050 L

$n'=Molarity\times V=0.250 M\times 0.050 L= 0.0125 mol$

Since 1 mole of Hcl reacts with 1 mole of NaoH. Then 0.0125 mole of HCl will react with 0.0125 mole of NaOH.

The enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

$q=\Delta H\times n=-57.2 kJ/mol \times 0.0125 mol= -0.715 kJ=-715 J$

q = heat released on reaction= -715 J

now, we calculate the heat gained by the solution.:

Q= -q = -(-715 J) = 715 J

m = mass of the solution = ?

Volume of the solution formed by mixing, v = 50.00 ml + 50.00 mL = 100.00 mL

Density of the solution = density of water = d = 1 g/mL

$mass=density\times volume=d\times v=1 g/ml \times 100.00 ml=100 g$

m = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{f}$ = final temperature = ?

$T_{i}$ = initial temperature = $19.50^oC$

$Q=mc\times (T_{f}-T_{i})$

Now put all the given values in the above formula, we get:

$715 J=100 g\times 4.18J/^oC\times (T_f-19.50)^oC$

$T_f=21.21 ^oC$

21.21°C will be the final temperature of the solution in a coffee cup calorimeter.

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koban [17]

Answer:

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mass = 3.2 mol * (58.3 g/mol)

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